"Y-system and functional dilogarithm identities"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
imported>Pythagoras0
imported>Pythagoras0
59번째 줄: 59번째 줄:
 
\end{align}
 
\end{align}
 
* {{수학노트|url=5항_관계식_(5-term_relation)}}
 
* {{수학노트|url=5항_관계식_(5-term_relation)}}
−
* [[5-term relation revisited from the viewpoint of Y-system]]
+
* [[Dilogarithm identity for CFT revisited]]
  
  

2013년 8월 23일 (금) 02:13 판

introduction


main results

  • Bloch group element

$$ \sum_{(\mathbf{i},u)\in S_{+}} Y_{\mathbf{i}}(u)\wedge (1+Y_{\mathbf{i}}(u))=0\in \Lambda^2 \mathbb{Q}(y)^{\times} $$ where $S_{+}=\{(\mathbf{i},u) |0\leq u \leq 2(h+h')-1,(\mathbf{i},u)\in P_{+}\}$.

  • functional dilogarithm identity

$$ \sum_{(\mathbf{i},u)\in S_{+}}L\left(\frac{Y_\mathbf{i}(u)}{1+Y_\mathbf{i}(u)}\right)=h r r' L(1) $$


bicoloring

  • what's $P_{+}$?
  • We give an alternate bicoloring on the pair of Dynkin diagrams. Let us fix bipartite decompositions of $I$ and $I'$.
  • Let $ \mathbf{I}= I\times I'$ and $\mathbf{I}=\mathbf{I}_{+}\sqcup \mathbf{I}_{-}$ where $\mathbf{I}_{+}=(I_{+}\times I'_{+}) \sqcup (I_{-}\times I'_{-})$ and $\mathbf{I}_{-}=(I_{+}\times I'_{-}) \sqcup (I_{-}\times I'_{+})$.
  • Let $\epsilon : \mathbf{I}\to \{1,-1\}$ be the function defined by $\epsilon(\mathbf{i})=\pm 1$ for $\mathbf{i}\in \mathbf{I}_{\pm}$ and $P_{\pm} =\{(\mathbf{i},u)\in \mathbf{I}\times\mathbb{Z}| \epsilon(\mathbf{i})(-1)^u=\pm 1\}$.
  • Roughly speaking, we want our alternate bicoloring interchanges their colors as $u\in \mathbb{Z}$ changes by 1.


an example

  • compute $\mathbb{Y}(A_2,A_1)$ explicitly
  • \(y_{m-1}y_{m+1}=y_m+1\)
  • Start with two variables \(y_1,y_2\).
  • \(y_3y_1=y_2+1\). so \(y_3=\frac{y_2+1}{y_1}\)
  • \(y_2y_4=y_3+1 \)implies \(y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2+1}{y_1y_2}\)
  • \(y_3y_5=y_4+1\) implies \(y_5=\frac{y_4+1}{y_3}= \frac{y_1+1}{y_2}\) we are getting Laurent polynomials
  • \(y_4y_6=y_5\) implies \(y_6=\frac{y_5+1}{y_4}= \frac{\frac{y_1+1}{y_2}+1}{\frac{y_1+y_2+1}{y_1y_2}}=\frac{y_1(y_1+1)+y_1y_2}{y_1+y_2+1}=y_1\)
  • rank 2 cluster algebra


observations

  • we saw that $$S=\left\{x,y,\frac{y+1}{x},\frac{x+y+1}{x y},\frac{x+1}{y}\right\}$$ forms a half-period of $\mathbb{Y}(A_2,A_1)$.
  • So we have $r=2,h=3$ and $r'=1,h'=2$.
  • They are all Laurent polynomials in $x$ and $y$.


dilogarithm identities

  • From this, one can get functional dilogarithm identities

\begin{align} &\sum_{a\in S}L\left(\frac{a}{1+a}\right) \notag \\ =& L\left(\frac{x}{1+x}\right)+L\left(\frac{y}{1+y}\right)+L\left(\frac{1+y}{x (1+\frac{1+y}{x})}\right)+L\left(\frac{1+x+y}{x y (1+\frac{1+x+y}{x y})}\right)+L\left(\frac{1+x}{(1+\frac{1+x}{y}) y}\right) \notag \\ =& L\left(\frac{x}{x+1}\right)+L\left(\frac{y}{y+1}\right)+L\left(\frac{y+1}{x+y+1}\right)+L\left(\frac{x+y+1}{x y+x+y+1}\right)+L\left(\frac{x+1}{x+y+1}\right) \notag \\ =&3L(1)=\frac{\pi^2}{2} \notag \end{align} and \begin{align} &\sum_{a\in S}L\left(\frac{1}{1+a}\right) \notag \\ =& L\left(\frac{1}{x+1}\right)+L\left(\frac{1}{y+1}\right)+L\left(\frac{1}{\frac{y+1}{x}+1}\right)+L\left(\frac{1}{\frac{x+y+1}{x y}+1}\right)+L\left(\frac{1}{\frac{x+1}{y}+1}\right) \notag \\ =& L\left(\frac{1}{x+1}\right)+L\left(\frac{1}{y+1}\right)+L\left(\frac{x}{x+y+1}\right)+L\left(\frac{x y}{x y+x+y+1}\right)+L\left(\frac{y}{x+y+1}\right) \notag \\ =&2L(1)=\frac{\pi^2}{3} \notag. \end{align}


history

 


related items


computational resource


expositions

 

 

articles

원본 주소 "https://wiki.mathnt.net/index.php?title=Y-system_and_functional_dilogarithm_identities&oldid=40171"